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3 years ago

Tell whether the ordered pair is a solution of the system of liner equation (3,2)x+y=5 y-2x=-4

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

7 0

Given:

The two linear equations are:

$x+y=5$

$y-2x=-4$

To find:

Whether the ordered pair (3,2) is a solution of the given system of liner equations.

Solution:

We have,

$x+y=5$ ...(i)

$y-2x=-4$ ...(ii)

If both the equations are satisfy by the point (3,2), then this point is the solution of the given system of equations.

Putting $x=3,y=2$ in (i), we get

$3+2=5$

$5=5$

This statement is true. It means the point (3,2) satisfies the equation (i).

Putting $x=3,y=2$ in (ii), we get

$2-2(3)=-4$

$2-6=-4$

$-4=-4$

This statement is true. It means the point (3,2) satisfies the equation (ii).

Since both the equations are satisfy by the point (3,2), therefore the point (3,2) is the solution of the given system of equations.

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Answer:

The MAD of {12, 10, 10, 8, 6, 7, 7, 12} is 2.

Step-by-step explanation:

Step 1 : Add all the values Up:

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Step 2 : Divide them by the number of items or numbers:

$\frac{12+10+10+8+6+7+7+12}{8}=9$

Step 3: Find the absolute deviations:

$12-9=3\\10-9=1\\10-9=1\\9-8=1\\9-6=3\\9-7=2\\9-7=2\\12-9=3\\$

Step 4: Add all the absolute deviations:

$3+1+1+1+3+2+2+3=16$

Step 5: Find the mean:

$\frac{3+1+1+1+3+2+2+3}{8}=2$

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45th percentile means that 45% of the data fall below that number while 55% (that is, 100-45) fall above it.

Now, arranging the number in ascending order;
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Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two coincident points

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by Pythagorean theorem we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by Zero product property we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0

3 years ago